Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,k,n) for(int i = (k); i < (n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) a.begin(), a.end()
#define MS(m,v) memset(m,v,sizeof(m))
#define D10 fixed<<setprecision(10)
typedef long long ll;
typedef vector<ll> vi;
typedef vector<string> vs;
typedef pair<int, int> pii;
typedef long double ld;
const int MOD = 1000000007;
const int INF = MOD + 1;
const ld EPS = 1e-10;
template<class T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<class T> T &chmax(T &a, const T &b) { return a = max(a, b); }
typedef int Weight;
struct Edge
{
	int from, to; Weight cost;
	bool operator < (const Edge& e) const { return cost < e.cost; }
	bool operator > (const Edge& e) const { return cost > e.cost; }
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
typedef vector<Weight> Array;
typedef vector<Array> Matrix;
void add_edge(Graph &g, int from, int to, Weight cost)
{
	g[from].push_back(Edge{ from, to, cost });
}

/*--------------------template--------------------*/
void dijkstra(Graph &g, Array &d, int s)
{
	d.assign(g.size(), INF);
	d[s] = 0;
	typedef pair<Weight, int> P;
	priority_queue<P, vector<P>, greater<P>> que;
	que.push(P(0, s));
	while (!que.empty())
	{
		Weight dist = que.top().first;
		int v = que.top().second;
		que.pop();
		if (d[v] < dist) continue;
		REP(i, g[v].size())
		{
			Edge e = g[v][i];
			if (d[e.to] > d[v] + e.cost)
			{
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
}


int main()
{
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	s--; t--;
	Graph g(n);
	REP(i, m)
	{
		int a, b;
		cin >> a >> b;
		a--; b--;
		add_edge(g, a, b, 1);
		add_edge(g, b, a, 1);
	}
	Array d1, d2;
	dijkstra(g, d1, s);
	dijkstra(g, d2, t);
	int dist = d1[t] - 1;
	vi v1(111111), v2(111111);
	REP(i, d1.size()) if (d1[i] != INF) v1[d1[i]]++;
	REP(i, d2.size()) if (d2[i] != INF) v2[d2[i]]++;
	ll ans = 0;
	REP(i, v1.size())
	{
		int t = dist - i - 1;
		if(t>=0&&t<111111) ans += v2[t]*v1[i];
	}
	cout << ans << endl;
	return 0;
}